4936 words

25 minutes

Analytic Functions and Power Series Theory

2025-12-10

Mathematic

Preparatory Work#

Complex Derivative#

Let $U\subset\mathbb C$ , $f:U\to\mathbb C$ , $z\in U$ . If there exists a $\mathbb C$ -linear function $df|_z\in\operatorname{Hom}_{\mathbb C}(\mathbb C,\mathbb C)$ such that for any $h\in\mathbb C$ , as $|h|\to 0$ , $|f(z+h)-f(z)-df|_z(h)|=o(|h|)$ , then $f$ is said to be complex differentiable at point $z$ , and $df|_z(1)$ is called its complex derivative, denoted by $f'(z)$ . Note that the complex derivative exists if and only if there exists $df|_z\in\operatorname{Hom}_{\mathbb C}(\mathbb C,\mathbb C)$ such that

\lim_{|h|\to 0}\left|\frac{f(z+h)-f(z)-df|_z(h)}{h}\right|=0,

i.e.,

\lim_{h\to 0}\frac{f(z+h)-f(z)-df|_z(h)}{h}=0,

or

\lim_{h\to 0}\frac{f(z+h)-f(z)}{h}=\lim_{h\to 0}\frac{df|_z(h)}{h}.

Since $\dim_{\mathbb C}\mathbb C=1$ , $df|_z(h)=f'(z)h$ , hence

f'(z)=\lim_{h\to 0}\frac{f(z+h)-f(z)}{h}.

Let $f(z)=u(\operatorname{Re}z,\operatorname{Im}z)+iv(\operatorname{Re}z,\operatorname{Im}z)$ . Set $z=x+iy$ , denote $\widetilde f(x,y):=f(x+iy)$ . Below we will not distinguish $\widetilde f$ and $f$ . If $h$ approaches $z$ from the real direction, then

f'(z)=\frac{\partial f}{\partial x}=\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x},

if $h$ approaches $z$ from the purely imaginary direction, then

f'(z)=\lim_{s\to 0}\frac{f(z+is)-f(z)}{is}=\frac{\partial u}{i\partial y}+i\frac{\partial v}{i\partial y}=-i\frac{\partial u}{\partial y}+\frac{\partial v}{\partial y}.

The equality of these two expressions gives a necessary condition for complex differentiability.

For $p=x+iy\in\mathbb C$ , formally define

\frac{\partial f}{\partial z}\bigg|_{p}=\frac{1}{2}\left(\frac{\partial f}{\partial x}\bigg|_{(x,y)}-i\frac{\partial f}{\partial y}\bigg|_{(x,y)}\right),\qquad\frac{\partial f}{\partial\overline z}\bigg|_{p}=\frac{1}{2}\left(\frac{\partial f}{\partial x}\bigg|_{(x,y)}+i\frac{\partial f}{\partial y}\bigg|_{(x,y)}\right),

and define $dz=dx+i\,dy,\,d\overline z=dx-i\,dy$ , then

\begin{aligned} \frac{\partial f}{\partial z}\,dz+\frac{\partial f}{\partial\overline z}\,d\overline z&=\frac{1}{2}\left(\frac{\partial f}{\partial x}-i\frac{\partial f}{\partial y}\right)(dx+i\,dy)+\frac{1}{2}\left(\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y}\right)(dx-i\,dy)\\ &=\frac{\partial f}{\partial x}\,dx+\frac{\partial f}{\partial y}\,dy=df. \end{aligned}

Theorem 1.1 $\quad$ Let $U\subset\mathbb C$ be an open set, $z\in U$ , then $f:U\to\mathbb C$ is differentiable at $z$ if and only if $\widetilde f$ is differentiable and

\frac{\partial f}{\partial\overline z}=0

i.e.,

\frac{\partial f}{\partial x}=-i\frac{\partial f}{\partial y}

i.e.,

\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y},\qquad\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}.

Proof $\quad$ Obviously if $f$ is complex differentiable at $z$ then $\widetilde{f}$ is real differentiable at that point, hence the necessity has been shown above.

Conversely, assume $\widetilde f$ is real differentiable and satisfies the Cauchy-Riemann equations. Then for $h\in\mathbb R^2$ , if $|h|\to 0$ , then $\widetilde f(z+h)-\widetilde{f}=A(h)+o(|h|)$ , where

A=\begin{pmatrix} \frac{\partial u}{\partial x}\big|_{z}&\frac{\partial u}{\partial y}\big|_{z}\\ \frac{\partial v}{\partial x}\big|_{z}&\frac{\partial v}{\partial y}\big|_{z} \end{pmatrix}.

Now we require $A(\cdot)$ to be $\mathbb C$ -linear. Note that the Cauchy-Riemann equations guarantee that $A$ is of the form $\begin{pmatrix} a&-b\\b&a \end{pmatrix}$ , so $A(\cdot)$ is exactly $(\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x})(\cdot)$ , thus $f$ is complex differentiable at $z$ . $\square$

A complex function is holomorphic if it is complex differentiable at every point of the entire $U$ .

Power Series#

Theorem 1.2 $\quad$ For every power series $\sum_{n=0}^{\infty}a_nz^n$ , there exists $R\in[0,\infty]$ , called the radius of convergence, such that

The series converges absolutely on $B(0,R)$ ;
The series converges uniformly on compact subsets of $B(0,R)$ ;
The series diverges on $\mathbb C\setminus\overline{B(0,R)}$ , and its terms are unbounded; d. On $B(0,R)$ , the sum of the series is analytic, its derivative is $\sum_{n=1}^{\infty}na_nz^{n-1}$ , and they have the same radius of convergence;
$1/R=\lim\limits_{n\to\infty}\sup\sqrt[n]{|a_n|}$ .

Proof $\quad$ If $|z|<R$ , then there exists $\rho\in(|z|,R)$ , so $\rho^{-1}>R^{-1}$ , hence $\exists N\in\mathbb N$ such that $\forall n\geq N$ , $|a_n|^{1/n}<\rho^{-1}$ , so $|a_n|<\rho^{-n}$ , hence $|a_nz^n|<(|z|/\rho)^n$ . Therefore

\sum_{n=N}^{\infty}\left|a_n\right||z^n|<\sum_{n=N}^{\infty}\left(\frac{|z|}{\rho}\right)^n<\infty,

so the power series converges absolutely at $z$ . For a given $\rho<R$ , choose $\rho_0\in(\rho,R)$ , there exists $N\in\mathbb N$ such that $\forall n\geq N$ , $|a_n|^{1/n}<\rho_0^{-1}$ , then similarly, for any $|z|\leq\rho$ , for any $\epsilon>0$ , there exists $\mathbb N'\geq N$ such that $\forall n,m\geq N'$ , we have

\sum_{j=n}^{m}\left|a_j\right||z^j|<\sum_{j=n}^{m}\left(\frac{\rho}{\rho_0}\right)^j<\epsilon,

so the power series converges uniformly on $\overline{B(0,\rho)}$ , hence it converges uniformly on compact subsets of $B(0,R)$ . If $|z|>R$ , then we can find $\rho\in(R,|z|)$ , so $\rho^{-1}<R^{-1}$ , then for any $N\in\mathbb N$ there exists $n>N$ such that $|a_n|>\rho^{-n}$ , so for infinitely many $n$ , $|a_nz^n|>(|z|/\rho)^n$ , thus the terms of the series are unbounded, and the series diverges.

Since $\sqrt[n]{n}\to 1$ , $\sum_{n=1}^{\infty}na_nz^{n-1}$ has the same radius of convergence. Finally, for $|z|<R$ , set

\begin{aligned} f(z)&=\sum_{n=0}^{\infty}a_nz^n=:s_n(z)+R_n(z),\\ s_n(z)&:=a_0+a_1z+\cdots+a_{n-1}z^{n-1},\qquad R_n(z):=\sum_{j=n}^{\infty}a_jz^j,\\ f_1(z)&:=\sum_{n=1}^{\infty}na_nz^{n-1}=\lim_{n\to\infty}s_n'(z). \end{aligned}

Now we prove $f'=f_1$ . For any $z_0\in B(0,R)$ , consider the identity

\begin{aligned} \frac{f(z)-f(z_0)}{z-z_0}-f_1(z_0)=\left(\frac{s_n(z)-s_n(z_0)}{z-z_0}-s_n'(z_0)\right)+(s_n'(z_0)-f_1(z_0))+\frac{R_n(z)-R_n(z_0)}{z-z_0}, \end{aligned}

where $z\neq z_0$ and $|z|,|z_0|<\rho<R$ . Note that

\begin{aligned} \frac{R_n(z)-R_n(z_0)}{z-z_0}&=\frac{1}{z-z_0}\sum_{j=n}^{\infty}(a_jz^j-a_jz_0^j)\\ &=\frac{1}{z-z_0}\sum_{j=n}^{\infty}(z-z_0)a_j(z^{j-1}+z^{j-2}z_0+\cdots+zz_0^{j-2}+z_0^{j-1})\\ &=\sum_{j=n}^{\infty}a_j(z^{j-1}+z^{j-2}z_0+\cdots+zz_0^{j-2}+z_0^{j-1}), \end{aligned}

thus as $n\to\infty$

\left|\frac{R_n(z)-R_n(z_0)}{z-z_0}\right|\leq\sum_{j=n}^{\infty}j|a_j|\rho^{j-1}\to 0.

On the other hand, $|s_n'(z_0)-f_1(z_0)|\to 0$ , so there exists $N\in\mathbb N$ such that for $n\geq N$ , each of the above two terms is less than $\epsilon/3$ . For any such $n$ , by the definition of derivative, there exists $\delta>0$ such that $\forall z\in B(z_0,\delta)$ we have

\left|\frac{s_n(z)-s_n(z_0)}{z-z_0}-s_n'(z_0)\right|<\frac{\epsilon}{3}.

Combining these we get

\left|\frac{f(z)-f(z_0)}{z-z_0}-f_1(z_0)\right|<\epsilon,

so $f_1(z_0)=f'(z_0)$ . $\square$

Cauchy Integral Theorem#

Line Integrals#

Let the equation of $\gamma:[a,b]\to\mathbb C$ be $z(t)$ . Define

\int_{\gamma}f\,dz:=\int_a^bf(z(t))z'(t)\,dt

and

\int_{\gamma}f\,d\overline{z}:=\overline{\int_{\gamma}\overline f\,dz}=\int_a^bf(z(t))\overline{z'(t)}\,dt.

For $dx=\frac{1}{2}(dz+d\overline{z}),\,dy=\frac{1}{2i}(dz-d\overline{z})$ , define

\begin{aligned} \int_{\gamma}f\,dx:=\frac{1}{2}\left(\int_{\gamma}f\,dz+\int_{\gamma}f\,d\overline{z}\right),\\ \int_{\gamma}f\,dy:=\frac{1}{2i}\left(\int_{\gamma}f\,dz-\int_{\gamma}f\,d\overline{z}\right), \end{aligned}

then

\begin{aligned} \int_{\gamma}f\,dz=\int_{\gamma}f\,dx+i\int_{\gamma}f\,dy,\qquad\int_{\gamma}f\,d\overline z=\int_{\gamma}f\,dx-i\int_{\gamma}f\,dy. \end{aligned}

Set $x:=\operatorname{Re}z,\,y:=\operatorname{Im}z,\,u:=\operatorname{Re}f,\,v:=\operatorname{Im}f$ , then

\begin{aligned} \begin{aligned} \int f\,dx&=\int_a^bf(z(t))x'(t)\,dt=\int_{\gamma}u\,dx+i\int_{\gamma}v\,dx,\\ \int f\,dy&=\int_a^bf(z(t))y'(t)\,dt=\int_{\gamma}u\,dy+i\int_{\gamma}v\,dy,\\ \int f\,dz&=\int_{\gamma}(u\,dx-v\,dy)+i\int_{\gamma}(v\,dx+u\,dy),\\ \int f\,d\overline z&=\int_{\gamma}(u\,dx+v\,dy)+i\int_{\gamma}(v\,dx-u\,dy).\\ \end{aligned} \end{aligned}

Define the integral with respect to arc length as

\int_{\gamma}f\,ds=\int_{\gamma}f\,|dz|:=\int_a^bf(z(t))|z'(t)|\,dt.

Rectifiable Arcs#

The length of an arc is defined as

l(\gamma):=\sup_{n,a=t_0<\cdots<t_n=b}\sum_{1}^n|z(t_j)-z(t_{j-1})|.

If $l(\gamma)<\infty$ , $\gamma$ is said to be rectifiable.

Theorem 2.3 $\quad$ An arc $\gamma:z=z(t)$ is rectifiable if and only if its real and imaginary parts are of bounded variation.

Proof $\quad$ Note that $|x(t_j)-x(t_{j-1})|\leq|z(t_j)-z(t_{j-1})|$ and $|y(t_j)-y(t_{j-1})|\leq|z(t_j)-z(t_{j-1})|$ , while $|z(t_j)-z(t_{j-1})|\leq|x(t_j)-x(t_{j-1})|+|y(t_j)-y(t_{j-1})|$ , thus if $x(t),y(t)$ are of bounded variation then $l(\gamma)<\infty$ , conversely, if $l(\gamma)<\infty$ then $x(t),y(t)$ are both of bounded variation. $\square$

Theorem 2.4 $\quad$ If $\gamma:[a,b]\to\mathbb C\xrightarrow{\sim}\mathbb R^2$ is $C^1$ , then $\gamma$ is rectifiable, and $l(\gamma)=\int_{\gamma}\,ds=\int_a^b|z'(t)|\,dt$ .

Proof $\quad$ Let $P$ be a partition of $[a,b]$ , then $|z(t_{j+1})-z(t_j)|=\left|\int_{t_j}^{t_{j+1}}z'(t)\,dt\right|\leq\int_{t_j}^{t_{j+1}}|z'(t)|\,dt$ , so

l_{P}(\gamma)=\sum_{j=0}^{n-1}|z(t_{j+1})-z(t_j)|\leq\sum_{j=0}^{n-1}\int_{t_j}^{t_{j+1}}|z'(t)|\,dt=\int_{a}^{b}|z'(t)|\,dt<\infty,

hence $l(\gamma)<\infty$ , $\gamma$ is rectifiable. Since $z'(t)$ is continuous on $[a,b]$ , it is uniformly continuous, so for any $\epsilon>0$ , there exists $\delta>0$ such that $\forall s,t\in[a,b],\,|t-s|<\delta$ , we have $|z'(t)-z'(s)|<\epsilon$ . Take a partition $P$ such that $\forall j,\,|t_j-t_{j-1}|<\delta$ . Then for any $t\in[t_j,t_{j+1}]$ , $|z'(t)-z'(t_{j})|<\epsilon$ . Thus

\begin{aligned} \int_{t_j}^{t_{j+1}}|z'(t)|\,dt&\leq\int_{t_j}^{t_{j+1}}|z'(t_j)|\,dt+\int_{t_j}^{t_{j+1}}|z'(t)-z'(t_j)|\,dt\\ &<\left|\int_{t_j}^{t_{j+1}}z'(t_j)\,dt\right|+(t_{j+1}-t_j)\epsilon\\ &\leq\left|\int_{t_j}^{t_{j+1}}z'(t)\,dt\right|+\left|\int_{t_j}^{t_{j+1}}(z'(t_j)-z'(t))\,dt\right|+(t_{j+1}-t_j)\epsilon\\ &<\left|\int_{t_j}^{t_{j+1}}z'(t)\,dt\right|+2(t_{j+1}-t_j)\epsilon\\ &=|z(t_{j+1})-z(t_j)|+2(t_{j+1}-t_j)\epsilon. \end{aligned}

Therefore

\begin{aligned} \int_{a}^{b}|z'(t)|\,dt&=\sum_{j=0}^{n-1}\int_{t_j}^{t_{j+1}}|z'(t)|\,dt\\ &<\sum_{j=0}^{n-1}\big(|z(t_{j+1})-z(t_j)|+2(t_{j+1}-t_j)\epsilon\big)\\ &=l_P(\gamma)+2(b-a)\epsilon<l(\gamma)+2(b-a)\epsilon. \end{aligned}

Since $\epsilon$ is arbitrary, letting $\epsilon\to 0$ yields $l(\gamma)=\int_{\gamma}\,ds=\int_a^b|z'(t)|\,dt$ . $\square$

If $\gamma$ is $C^1$ and $f(z)$ is continuous on $\gamma$ , then $f(z(t))|z'(t)|$ is uniformly continuous.

Cauchy Integral Theorem#

After decomposing mappings into real and imaginary parts, theorems concerning exact forms, closed forms, and curve integrals can be directly applied. If $p,q$ are respectively the partial derivatives of some complex function $U$ with respect to $x$ and $y$ , then the integral $\int_{\gamma}(p\,dx+q\,dy)$ depends only on the endpoints. Indeed, let $p=a+ib,q=c+id$ , then

p\,dx+q\,dy=(a+ib)\,dx+(c+id)\,dy=(a\,dx+c\,dy)+i(b\,dx+d\,dy),

if we can find $V,W$ such that $dV=a\,dx+c\,dy,dW=b\,dx+d\,dy$ , then $U:=V+iW$ satisfies $dU=p\,dx+q\,dy$ . Conversely, if there exists a complex-valued function $U$ such that $dU=p\,dx+q\,dy$ , then set $U=V+iW$ , we have

\frac{\partial U}{\partial x}=a\,dx+ib\,dx=\frac{\partial V}{\partial x}+i\frac{\partial W}{\partial x},

giving $\partial V/\partial x=a,\,\partial W/\partial x=b$ , similarly, $\partial V/\partial y=c,\,\partial W/\partial y=c$ , so $a\,dx+c\,dy,\,b\,dx+d\,dy$ are each exact, hence the integral $\int_\gamma(p\,dx+q\,dy)$ is independent of the path.

If the integral $\int_{\gamma}f\,dz$ is independent of the specific path, i.e., $f\,dz$ is exact, then there exists $F(z)$ such that $dF=f\,dx+if\,dy$ , so

\frac{\partial F}{\partial x}=f,\qquad\frac{\partial F}{\partial y}=if,

thus $F(z)$ satisfies the Cauchy-Riemann equations.

From the above discussions, it can already be concluded that since $dz$ , $z\,dz$ are exact, or more generally, $z^n\,dz\,(n\geq 0)$ are all exact, as long as the integral is closed, we have

\oint dz=\oint z\,dz=\oint z^n\,dz=0.

Secondly, if the region is simply connected, $f$ is holomorphic and its partial derivatives are continuous, then $\oint f\,dz=0$ . This is because by the Cauchy-Riemann equations, we have respectively

\begin{aligned} d(u\,dx-v\,dy)&=-\left(\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}\right)dx\wedge dy=0,\\ d(v\,dx+u\,dy)&=\left(\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y}\right)dx\wedge dy=0, \end{aligned}

so $f\,dz$ is exact, yielding $\oint f\,dz=0$ .

This is the proof initially given by Cauchy. In the corollary of Poincaré’s lemma, if $U$ is simply connected, then closed is equivalent to exact on $U$ , but the proof of Poincaré’s lemma requires the partial derivatives of $f$ to be continuous. Later, Goursat found that the condition of continuous partial derivatives is unnecessary, thus giving the following first version of the Cauchy integral theorem.

Theorem 2.5 (Goursat) $\quad$ Let $R$ be a rectangle, $f$ holomorphic on $\overline{R}$ , then

\int_{\partial R}f\,dz=0.

Proof $\quad$ For a region $H$ , denote $\eta(H):=\int_{\partial H}f\,dz$ . For $R$ , divide it into four congruent rectangles $R_1^1,R_2^1,R_3^1,R_4^1$ , then

\eta(R)=\eta(R_1)+\eta(R_2)+\eta(R_3)+\eta(R_4).

Thus at least one $R_{j_1}^1$ satisfies $|\eta(R_j^1)|\geq 4^{-1}|\eta(R)|$ . Continue dividing this $R_{j_1}^1$ , there exists $R_{j_2}^2$ satisfying $|\eta(R_{j_2}^2)|\geq 4^{-1}|\eta(R_{j_2}^2)|$ . Continuing this way, we obtain a descending sequence of rectangles $\{R_{j_k}^k\}_{k=1}^{\infty}$ , where $|\eta(R_{j_k}^k)|\geq 4^{-k}|\eta(R)|$ . By the Cantor intersection theorem, $\bigcap_{k=1}^{\infty}R_{j_k}^k=z_0\in\mathbb R$ . Since $R$ is complex differentiable at $z_0$ , for any $\epsilon>0$ , there exists $\delta>0$ such that

|f(z)-f(z_0)-f'(z_0)(z-z_0)|<\epsilon|z-z_0|,\qquad\forall z\in B(z_0,\delta).

Take $n$ sufficiently large such that $R_{j_n}^n\subset B(z_0,\delta)$ , let $R_n:=R_{j_n}^n$ . From previous discussions, we know that integrals of $dz$ and $z\,dz$ are zero, so

\begin{aligned} |\eta(R_n)|&=\left|\int_{\partial R_n}(f(z)-f(z_0)-f'(z_0)(z-z_0))\,dz\right|\\ &\leq\int_{\partial R_n}|f(z)-f(z_0)-f'(z_0)(z-z_0)|\,|dz|<\epsilon\int_{\partial R_n}|z-z_0|\,ds. \end{aligned}

Let the diagonal of the original rectangle be $d$ , perimeter be $L$ , then the diagonal of $R_n$ is $d_n=2^{-n}d$ , perimeter $L_n=2^{-n}L$ , so

4^{-n}|\eta(R)|\leq|\eta(R_n)|<\epsilon\int_{\partial R_n}d_n\,ds=\epsilon d_nL_n=4^{-n}\epsilon dL,

i.e., $|\eta(R)|<\epsilon dL$ . Since $\epsilon$ is arbitrary, we get $\int_{\partial R}f\,dz=\eta(R)=0$ . $\square$

Theorem 2.6 $\quad$ Let $R$ be a rectangle, with finitely many points $\{\zeta_j\}_1^n$ in its interior, $f$ holomorphic on $\overline R\setminus\{\zeta_j\}$ . If

\lim_{z\to\zeta_j}(z-\zeta_j)f(z)=0,\qquad\forall j=1,2,\cdots,n,

then

\int_{\partial R}f\,dz=0.

Proof $\quad$ Decompose $R$ into small rectangles so that each contains at most one $\zeta_j$ . Thus the problem reduces to the case of a single point set $\{\zeta_j\}=\{\zeta\}$ . In this case, divide $R$ into nine small rectangles, with only the central one $R_0$ containing $\zeta$ . Apply the first version of Cauchy’s integral theorem to the other eight rectangles, then sum all integrals to get

\int_{\partial R}f\,dz=\int_{\partial R_0}f\,dz.

For any $\epsilon$ , by the theorem condition, we can choose $R_0$ sufficiently small such that on $\partial R$ , $|f(z)|<\epsilon/|z-\zeta|$ . Since such $R_0$ can be chosen arbitrarily, assume $R_0$ is a square with $\zeta$ at its center. Then

\left|\int_{\partial R_0}f\,dz\right|\leq\epsilon\int\frac{|dz|}{|z-\zeta|}.

Now estimate the integral on the right. The integral is symmetric, so we only need to compute one side. Suppose $R_0$ has side length $2s$ , after translation $\zeta=0$ , consider the top side of the rectangle, then $|dz|=dx$ , $|z|\geq s$ . So the integral on the top side is less than $\int_{-s}^s\frac{dx}{s}=2$ , overall

\int_{\partial R_0}\frac{|dz|}{|z-\zeta|}<8,\qquad \left|\int_{\partial R_0}f\,dz\right|<8\epsilon.

Since $\epsilon$ is arbitrary, the proposition follows. $\square$

With the Cauchy integral theorem for rectangles, we can prove the Cauchy integral theorem for disks without requiring the partial derivatives of $f$ to be continuous.

Theorem 2.7 $\quad$ Let $f$ be holomorphic on an open disk $D$ , then for every closed curve $\gamma$ in $D$ , we have

\int_{\gamma}f\,dz=0.

Proof $\quad$ Let the center be $x_0+iy_0$ . For $x+iy\in D$ , define $\sigma$ as the path from $x_0+iy_0$ horizontally to $x+iy_0$ , then vertically to $x+iy$ . By Goursat’s theorem, if $\tau$ is the path first vertically then horizontally, then the integrals of $f$ along these two paths are equal. Set

F(z)=\int_{\sigma}f\,dz=\int_{\tau}f\,dz,

considering the integrals along $\tau$ and $\sigma$ with respect to partial $x$ and partial $y$ respectively, we find

\frac{\partial F}{\partial x}\bigg|_{z}=f(z),\qquad\frac{\partial F}{\partial y}\bigg|_z=if(z),

so $f\,dz$ is exact, hence the integral of $f$ over any closed curve is zero. $\square$

Corollary 2.8 $\quad$ Let $D$ be an open disk with finitely many points $\{\zeta_j\}_1^n$ in its interior. Let $f$ be holomorphic on $D$ , and

\lim_{z\to\zeta_j}(z-\zeta_j)f(z)=0,\qquad\forall j=1,2,\cdots,n,

then for every closed curve $\gamma$ in $D$ , we have

\int_{\gamma}f\,dz=0.

Proof $\quad$ Replace the Goursat theorem used in the proof of the previous theorem with its version with punctured points. $\square$

Winding Number#

First consider the winding number on $S^1$ . In 3.1 to 3.6, we assume curves are $C^1$ .

Definition 3.1 $\quad$ Let $S^1\subset\mathbb R^2$ be the unit circle and $c:I=[0,b]\to S^1,t\mapsto(x(t),y(t))$ a closed curve. Find $\varphi_0$ such that $\cos(\varphi_0)=x(0),\sin(\varphi_0)=y(0)$ . Define

\begin{aligned} \varphi:I&\longrightarrow\mathbb R,\\ t&\longmapsto\varphi(t):=\varphi_0+\int_0^t(xy'-yx')\,d\tau, \end{aligned}

called the angle function of the closed curve $c$ .

The geometric meaning of the angle function is given immediately by the following proposition:

Lemma 3.2 $\quad$ Let $c:[0,b]\to S^1$ be a closed curve, $\varphi$ its angle function, then

\begin{cases} \cos(\varphi(t))=x(t),\\ \sin(\varphi(t))=y(t). \end{cases}

Proof $\quad$ Set $F(t):=(x(t)-\cos(\varphi(t)))^2+(y(t)-\sin(\varphi(t)))^2$ , then

\begin{aligned} F&=(x-\cos\varphi)^2+(y-\sin\varphi)^2\\ &=x^2-2x\cos\varphi+\cos^2\varphi+y^2-2y\sin\varphi+\sin^2\varphi. \end{aligned}

Note that $\varphi'=xy'-yx'$ , differentiating $F$ gives

\begin{aligned} F'&=2xx'-2(x'\cos\varphi-x\varphi'\sin\varphi)-2\varphi'\cos\varphi\sin\varphi\\ &\quad+2yy'-2(y'\sin\varphi+y\varphi'\cos\varphi)+2\varphi'\cos\varphi\sin\varphi\\ &=2(xx'+yy')+2(-x'-y\varphi')\cos\varphi+2(-y'+x\varphi')\sin\varphi. \end{aligned}

Since $x(t)^2+y(t)^2=1$ , we have $2x(t)x'(t)+y(t)y'(t)=0$ , i.e., $xx'+yy'=0$ . $x,y$ or $x',y'$ obviously cannot be identically zero, the only possibility is that there exists $k(t)$ such that $x'=-ky,\,y'=kx$ . Thus

\varphi'(t)=xy'-yx'=k(x^2+y^2)=k(t),

so $x'=-\varphi'y,\,y'=\varphi'x$ . Substituting yields

F'\equiv 0.

This shows $F$ is constant. Note that at $t=0$

F=F(0)=(x(0)-\cos(\varphi_0))^2+(y(0)-\sin(\varphi_0))^2=0,

so $F$ is identically zero on $[0,b]$ . This proves $\cos(\varphi(t))=x(t),\sin(\varphi(t))=y(t)$ . $\square$

Definition 3.3 $\quad$ Let $c:[0,b]\to S^1$ be a closed curve, $\varphi$ its angle function. Define

n(c):=\frac{1}{2\pi}(\varphi(b)-\varphi(0)),

called the winding number of $c$ . Since $c$ is closed, by Lemma 3.2, $n(c)\in\mathbb Z$ .

Now we extend the concept of winding number to general closed curves.

Definition 3.4 $\quad$ Let $p\in\mathbb R^2$ , $\gamma:[0,b]\to\mathbb R^2\setminus\{p\}$ , set $d(t):=\gamma(t)-p,\,\rho(t):=|d(t)|,\,c(t):=d(t)/\rho(t)$ , then $d,\rho,c\in C^1$ , and $\gamma(t)=p+\rho(t)c(t)$ . Define

n(\gamma,p):=n(c)

as the winding number of $\gamma$ about $p$ .

Proposition 3.5 $\quad$ Let $\gamma(t):=p+\rho(t)c(t)$ be a closed curve $[0,b]\to\mathbb R^2\setminus\{p\}$ . Set $(x(t),y(t)):=\gamma(t)-p$ , then

n(\gamma,p)=\frac{1}{2\pi}\int_{\gamma}\omega_0,\qquad\omega_0:=-\frac{y}{x^2+y^2}\,dx+\frac{x}{x^2+y^2}\,dy.

Proof $\quad$ Note that $c(t)=(x(t)/\rho(t),y(t)/\rho(t))$ , thus

\begin{aligned} n(c)&=\frac{1}{2\pi}\int_0^b\left(\frac{x(t)}{\rho(t)}\left(\frac{y(t)}{\rho(t)}\right)'-\frac{y(t)}{\rho(t)}\left(\frac{x(t)}{\rho(t)}\right)'\right)\,dt\\ &=\frac{1}{2\pi}\int_0^b\left(\frac{x}{\rho}\frac{y'\rho-y\rho'}{\rho^2}-\frac{y}{\rho}\frac{x'\rho-x\rho'}{\rho^2}\right)\,dt=\frac{1}{2\pi}\int_0^{b}\frac{xy'-yx'}{\rho^2}\,dt. \end{aligned}

On the other hand

\begin{aligned} \frac{1}{2\pi}\int_{\gamma}\omega_0&=\frac{1}{2\pi}\int_{\gamma}\left(-\frac{y}{x^2+y^2}\,dx+\frac{x}{x^2+y^2}\,dy\right)\\ &=\frac{1}{2\pi}\int_0^b\left(-\frac{yx'}{x^2+y^2}+\frac{xy'}{x^2+y^2}\right)\,dt=\frac{1}{2\pi}\int_0^b\frac{xy'-yx'}{\rho^2}\,dt,\\ \end{aligned}

so $n(\gamma,p)=n(c)=\frac{1}{2\pi}\int_{\gamma}\omega_0$ . $\square$

Theorem 3.6 $\quad$ Let $\gamma_0,\gamma_1:[0,b]\to\mathbb R^2\setminus\{p\}$ be two closed curves. Then $\gamma_0$ and $\gamma_1$ are freely homotopic if and only if $n(\gamma_0,p)=n(\gamma_1,p)$ .

Proof $\quad$ Take the exterior derivative of the $1$ -form $\omega_0$ in Proposition 3.5:

\begin{aligned} d\omega_0&=d\left(-\frac{y}{x^2+y^2}\,dx+\frac{x}{x^2+y^2}\,dy\right)\\ &=d\left(\frac{-y}{x^2+y^2}\right)\wedge dx+d\left(\frac{x}{x^2+y^2}\right)\wedge dy\\ &=\frac{-x^2+y^2}{(x^2+y^2)^2}\,dy\wedge dx+\frac{-x^2+y^2}{(x^2+y^2)^2}\,dx\wedge dy=0, \end{aligned}

so $\omega_0$ is closed. Integrals of closed forms along freely homotopic paths are always equal, so by Proposition 3.5, necessity is proven.

Sufficiency: Let $\gamma_j=p+\rho_jc_j$ , let $\varphi_j$ be the angle function of $c_j$ , set $\varphi(t,\tau):=(1-\tau)\varphi_0(t)+\tau\varphi_1(t)$ , $H(t,\tau):=p+(\cos(\varphi(t,\tau)),\sin(\varphi(t,\tau)))$ , and set $\widehat{\gamma}_j:=p+c_j$ and

H_j(t,\tau):=p+\frac{\rho_j(t)}{(1-\tau)+\tau\rho_j(t)}c_j(t).

Obviously $H_j$ is a free homotopy from $\gamma_j$ to $\widehat{\gamma}_j$ . If $n(\gamma_0,p)=n(\gamma_1,p)$ , then $\forall\tau\in[0,1]$ we have

\begin{aligned} \varphi(b,\tau)-\varphi(0,\tau)&=(1-\tau)\varphi_0(b)+\tau\varphi_1(b)-\big((1-\tau)\varphi_0(0)+\tau\varphi_1(0)\big)\\ &=(1-\tau)(\varphi_0(b)-\varphi_0(0))+\tau(\varphi_1(b)-\varphi_1(0))\\ &=(1-\tau)2\pi n(\gamma_0,p)+\tau 2\pi n(\gamma_1,p)=2\pi n(\gamma_0,p), \end{aligned}

so $H$ is a free homotopy from $\widehat{\gamma}_0$ to $\widehat{\gamma}_1$ . In summary

\gamma_0\simeq\widehat{\gamma}_0\simeq\widehat{\gamma}_1\simeq\gamma_1,

proving sufficiency. $\square$

Generally, let $\gamma:[a,b]\to\mathbb C$ be a closed rectifiable curve, $w\notin\operatorname{im}(\gamma)$ , define

\begin{aligned} u:[a,b]&\longrightarrow S^1=\partial\mathbb D\subset\mathbb C\\ t&\longmapsto\frac{\gamma(t)}{|\gamma(t)-w|}, \end{aligned}

then by Theorem 3.6, if $\gamma$ is once differentiable then $n(\gamma,w)=n(u)$ . Now we extend the winding number to rectifiable curves and express it using complex integrals.

Lemma 3.7 (Special case of Lifting Lemma) $\quad$ Let $\pi:\mathbb R\to S^1,s\mapsto e^{i2\pi s}$ be the universal cover. Then there exists a lift $\tilde u:[a,b]\to\mathbb R$ of $u$ such that $\pi\circ\tilde{u}=u$ .

Proof $\quad$ If $u$ is once differentiable then $\tilde u$ can be taken as $\varphi(t)/2\pi$ , which also proves existence for piecewise differentiable curves. Since rectifiable curves can be approximated by piecewise differentiable curves, the lemma follows. $\square$

Definition 3.8 $\quad$ Let $\gamma:[a,b]\to\mathbb C$ , $w\notin\operatorname{im}(\gamma)$ , $u$ defined as above, $\tilde u$ its lift. Define the winding number of $\gamma$ about $w$ as $n(\gamma,w)=\tilde u(b)-\tilde u(a)$ , an extension of Definition 3.3.

Theorem 3.9 $\quad$ The winding number varies continuously under free homotopy, i.e., let $\gamma_j:[a,b]\to\mathbb C$ , $w\notin\operatorname{im}(\gamma_j)$ rectifiable, $H(t,\tau)$ a free homotopy from $\gamma_1$ to $\gamma_2$ , then $n(\tau):=n(H(\cdot,\tau),w)$ is continuous. In particular, if $\gamma_1,\gamma_2$ are freely homotopic, then $n(\gamma_1,w)=n(\gamma_2,w)$ .

Proof $\quad$ Apply the lifting lemma to $H$ to get $\tilde H$ , then $n(\tau)=\tilde H(b,\tau)-\tilde H(a,\tau)$ . Since $\tilde H$ is continuous, $n(\tau)$ is naturally continuous. Since $n(\tau)$ always takes integer values, we have $n(\gamma_1,w)=n(0)=n(1)=n(\gamma_2,w)$ . $\square$

Theorem 3.10 $\quad$ Let $\gamma:[a,b]\to\mathbb C$ be a rectifiable curve, $w\notin\operatorname{im}(\gamma)$ . Then

n(\gamma,w)=\frac{1}{2\pi i}\int_{\gamma}\frac{1}{z-w}\,dz.

Remark $\quad$ Rectifiable curves can be uniformly approximated by piecewise differentiable curves, so it suffices to prove the proposition for the $C^1$ case. Two proofs are given here.

Proof 1 $\quad$ Let $\zeta(t):=\gamma(t)-w=x(t)+iy(t)$ , then

\begin{aligned} \frac{d\zeta}{\zeta}&=\frac{dx+i\,dy}{x+iy}=\frac{(dx+i\,dy)(x-iy)}{(x+iy)(x-iy)}\\ &=\frac{x\,dx+y\,dy}{x^2+y^2}+i\left(\frac{x\,dy-y\,dx}{x^2+y^2}\right)\\ &=\frac{x\,dx+y\,dy}{x^2+y^2}+i\omega_0, \end{aligned}

hence

\begin{aligned} \frac{1}{2\pi i}\int_{\gamma}\frac{1}{z-w}\,dz&=\frac{1}{2\pi i}\int_{\gamma}\frac{d\zeta}{\zeta}\\ &=\frac{1}{2\pi i}\int_{\gamma}i\omega_0+\frac{1}{2\pi i}\int_{\gamma}\frac{x\,dx+y\,dy}{x^2+y^2}\\ &=n(\gamma,w)+\frac{1}{2\pi i}\int_{\gamma}\frac{x\,dx+y\,dy}{x^2+y^2}. \end{aligned}

Note that

\frac{x\,dx+y\,dy}{x^2+y^2}=\frac{1}{2}\frac{d(x^2+y^2)}{x^2+y^2}=\frac{1}{2}d\log(x^2+y^2)

is exact, so its integral around $\gamma$ is zero. Therefore

n(\gamma,w)=\frac{1}{2\pi i}\int_{\gamma}\frac{1}{z-w}\,dz.

Proof 2 $\quad$ Set

h(t):=\frac{1}{2\pi i}\int_a^t\frac{\gamma'(s)}{\gamma(s)-w}\,ds,\qquad g(t):=e^{-2\pi ih(t)}(\gamma(t)-w),

then

\begin{aligned} h'(t)&=\frac{1}{2\pi i}\frac{\gamma'(t)}{\gamma(t)-w},\qquad g'(t)\equiv 0, \end{aligned}

so $g(t)$ is constant; further $g=g(a)=(\gamma(a)-w)$ . Thus

e^{2\pi ih(t)}=\frac{\gamma(t)-w}{\gamma(a)-w}.

On the other hand

e^{2\pi i\tilde u(t)}=\frac{\gamma(t)-w}{|\gamma(t)-w|},

set $v(t):=\log|\gamma(t)-w|$ , then $v(b)=v(a)$ and

e^{2\pi i\tilde u(t)+v(t)}=\gamma(t)-w,

so there exists a constant $\alpha\in\mathbb C$ such that

2\pi ih(t)=2\pi i\tilde u(t)+v(t)+\alpha,

hence

n(\gamma,w)=\tilde u(b)-\tilde u(a)=h(b)-h(a)=\frac{1}{2\pi i}\int_{\gamma}\frac{1}{z-w}\,dz.

$\square$

Cauchy Integral Formula#

This section proves that holomorphic functions are smooth, and along the way yields important theorems such as the Fundamental Theorem of Algebra.

Basic Statement#

Let $f(z)$ be holomorphic on an open disk, $\gamma$ a closed curve inside it. For a point $a\in\Delta$ outside $\gamma$ , consider the function

F(z):=\frac{f(z)-f(a)}{z-a},

this function is holomorphic on $D\setminus\{a\}$ , so as $z\to a$ , $(z-a)F(z)=f(z)-f(a)\to 0$ . Then applying the Cauchy integral theorem gives

\int_{\gamma}\frac{f(z)-f(a)}{z-a}\,dz=0,

i.e.,

\int_{\gamma}\frac{f(z)}{z-a}\,dz=f(a)\int_{\gamma}\frac{dz}{z-a}.

Note that the integral on the right is precisely $2\pi i\cdot n(\gamma,a)$ . Generally, we have the following theorem.

Theorem 4.1 $\quad$ Let $f$ be holomorphic on an open disk $D$ , $\gamma$ a closed curve in $D$ , $a\in\mathbb C\setminus\operatorname{im}(\gamma)$ , then

n(\gamma,a)f(a)=\frac{1}{2\pi i}\int_{\gamma}\frac{f(z)}{z-a}\,dz.

Proof $\quad$ The case $a\in D$ has been argued above. If $a\notin D$ , then the winding number $n(\gamma,a)$ and the integral on the right are also zero, so the formula still holds. $\square$

For a fixed closed curve $\gamma$ , for those points $z$ with winding number $n(\gamma,z)=1$ , we have

f(z)=\frac{1}{2\pi i}\int_{\gamma}\frac{f(w)}{w-z}\,dw,\qquad n(\gamma,z)=1.

This is the Cauchy integral formula. In particular, for an open disk $D\subset\mathbb C$ , let $f:\overline{D}\to\mathbb C$ be holomorphic, then taking a slightly larger open disk on which $f$ is also holomorphic yields

f(z)=\frac{1}{2\pi i}\int_{\partial D}\frac{f(w)}{w-z}\,dw.

Higher Order Derivatives#

Lemma 4.2 $\quad$ Let $\phi(w)$ be a continuous function on an arc $\gamma$ , then the function

\Phi_n(z):=\int_{\gamma}\frac{\phi(w)}{(w-z)^n}\,dw

is analytic on $\mathbb C\setminus\operatorname{im}(\gamma)$ , and $\Phi_n'(z)=n\Phi_{n+1}(z)$ .

Proof $\quad$ First prove $\Phi_1$ is continuous. Let $z_0$ be a point not on $\gamma$ , take $B(z_0,\delta)$ not intersecting $\gamma$ , then for all $z\in B(z_0,\delta/2),\,w\in\gamma$ , we have $|w-z|>\delta/2$ . Thus from

\frac{1}{w-z}-\frac{1}{w-z_0}=\frac{z-z_0}{(w-z)(w-z_0)}

we get

\begin{aligned} |\Phi_1(z)-\Phi_1(z_0)|&=\left|(z-z_0)\int_{\gamma}\frac{\phi(w)\,dw}{(w-z)(w-z_0)}\right|\\ &<\left|z-z_0\right|\frac{4}{\delta^2}\int_{\gamma}|\phi(w)|\,|dw|=:C\left|z-z_0\right|, \end{aligned}

where $C$ is a constant. This proves $\Phi_1$ is continuous at any $z_0$ not on $\gamma$ . Generally, from

\frac{1}{(w-z)^n}-\frac{1}{(w-z_0)^n}=\frac{1}{(w-z)^{n-1}(w-z_0)}+\frac{(z-z_0)}{(w-z)^n(w-z_0)}-\frac{1}{(w-z_0)^n}

we get

\begin{aligned} \Phi_n(z)-\Phi_n(z_0)&=\left[\int_{\gamma}\frac{\phi(w)\,dw}{(w-z)^{n-1}(w-z_0)}-\int_{\gamma}\frac{\phi(w)\,dw}{(w-z_0)^n}\right]\\ &\quad+(z-z_0)\int_{\gamma}\frac{\phi(w)\,dw}{(w-z)^n(w-z_0)}. \end{aligned}

Now prove by induction that $\Phi_n$ is continuous. Assume the proposition holds for $n-1$ , then taking $\psi:=\phi(w)/(w-z_0)$ and applying the induction hypothesis shows the function

\Psi_{n-1}(z_0;z):=\int_{\gamma}\frac{\phi(w)\,dw}{(w-z)^{n-1}(w-z_0)}

is continuous, so as $z\to z_0$ ,

\int_{\gamma}\frac{\phi(w)\,dw}{(w-z)^{n-1}(w-z_0)}-\int_{\gamma}\frac{\phi(w)\,dw}{(w-z_0)^n}\;\to\;\int_{\gamma}\frac{\phi(w)\,dw}{(w-z_0)^n}-\int_{\gamma}\frac{\phi(w)\,dw}{(w-z_0)^n}=0,

and the remaining term can be shown to tend to zero as $z\to z_0$ by taking $\delta/2$ similarly. This proves $F_n$ is continuous.

For $n=1$ , note that

\frac{\Phi_1(z)-\Phi_1(z_0)}{z-z_0}=\int_{\gamma}\frac{\phi(w)\,dw}{(w-z)(w-z_0)},

we have already shown the integral on the right is continuous in $z$ , so as $z\to z_0$ we get

\Phi_1'(z_0)=\int_{\gamma}\frac{\phi(w)\,dw}{(w-z_0)^2}=F_2(z_0),\qquad\forall z_0\notin\operatorname{im}(\gamma).

Now prove by induction that $\Phi_n'(z)=n\Phi_{n+1}(z)$ . Assume the proposition holds for $n-1$ , then

\begin{aligned} \frac{\Phi_n(z)-\Phi_n(z_0)}{z-z_0}&=\frac{1}{z-z_0}\left[\int_{\gamma}\frac{\phi(w)\,dw}{(w-z)^{n-1}(w-z_0)}-\int_{\gamma}\frac{\phi(w)\,dw}{(w-z_0)^n}\right]\\ &\quad+\int_{\gamma}\frac{\phi(w)\,dw}{(w-z)^n(w-z_0)}\\ &=\frac{\Psi_{n-1}(z_0;z)-\Psi_{n-1}(z_0;z_0)}{z-z_0}+\int_{\gamma}\frac{\phi(w)\,dw}{(w-z)^n(w-z_0)}, \end{aligned}

taking $z\to z_0$ and using the induction hypothesis gives

\Phi_n'(z_0)=(n-1)\Psi_{n}(z_0;z_0)+\Phi_{n+1}(z_0).

But $\Psi_k(z_0;z_0)=\Phi_{k+1}(z_0)$ , so

\Phi_n'(z_0)=n\Phi_{n+1}(z_0),\qquad\forall z_0\notin\operatorname{im}(\gamma).

$\square$

Proposition 4.3 $\quad$ Let $D\subset\mathbb C$ be an open disk, $f$ holomorphic on $\overline D$ , then for any $z\in\mathbb D$ , $f$ is $C^{\infty}$ at point $z$ , and

f^{(n)}(z)=\frac{n!}{2\pi i}\int_{\partial D}\frac{f(\zeta)\,d\zeta}{(\zeta-z)^{n+1}}.

Proof $\quad$ Apply the Cauchy integral theorem. $\square$

General Form of Cauchy Integral Theorem#

Definition 5.1 (Chain, Cycle) $\quad$ Let $U\subset\mathbb C$ be an open set, $\mathcal R$ the set of all rectifiable curves on $U$ , denote $C_0(U):=\mathbb Z^{\oplus U},C_1(U):=\mathbb Z^{\oplus\mathcal R}$ , abbreviated as $C_0,C_1$ . A 1-chain is a formal sum $\gamma=\sum_{j=1}^na_j\gamma_j\in C_1$ . Denote $\operatorname{im}(\gamma):=\bigcup_1^n\operatorname{im}(\gamma_j)$ . For any $f\in C(\operatorname{im}(\gamma))$ , define

\int_{\gamma}f\,dz:=\sum_{j=1}^na_j\int_{\gamma_j}f\,dz.

Let $\iota:U\to C_0$ be the canonical inclusion map. Define

\begin{aligned} \partial:C_1&\longrightarrow C_0\\ \sum_1^na_j\gamma_j&\longmapsto\sum_1^na_j(\iota(\gamma_j(\beta_j))-\iota(\gamma_j(\alpha_j))), \end{aligned}

where $\gamma_j:[\alpha_j,\beta_j]\to\mathbb C$ . Elements in $\ker(\partial)$ are called cycles. For a cycle $\gamma$ and $z\notin\operatorname{im}(\gamma)$ , define

n(\gamma,z):=\frac{1}{2\pi i}\int_{\gamma}\frac{dw}{w-z}.

Definition 5.2 $\quad$ Let $U\subset\mathbb C$ be an open set, $\gamma\in C_1(U)$ a cycle. $\gamma$ is said to be homologous to $0$ in $U$ if $\forall z\in\mathbb C\setminus U$ , $n(\gamma,z)=0$ .

In the following, we may assume the cycle $\gamma$ is a sum of closed curves.

Lemma 5.3 $\quad$ Let $U\subset\mathbb C$ be a simply connected open set, then any cycle on $U$ is homologous to $0$ .

Proof $\quad$ Each $\gamma_j$ is homotopic to a point, and homotopy preserves winding number. $\square$

Theorem 5.4 (General Form of Cauchy Integral Theorem) $\quad$ Let $U\subset\mathbb C$ be an open set, $\gamma$ a cycle on $U$ homologous to $0$ . Let $f$ be a holomorphic function on $U$ . Then

\int_{\gamma}f\,dz=0.

Proof $\quad$ First assume $U$ is bounded. For any $\delta>0$ , take $n\in\mathbb Z$ such that $2^{-n}<\delta$ , consider the family of closed squares $\mathcal Q_n$ of side length $2^{-n}$ in the complex plane. Let $\{Q_j\}_{j\in J}\subset\mathcal Q_n$ be those closed squares entirely contained in $U$ , since $U$ is bounded, it is indeed a finite set. Choose $\delta$ sufficiently small so that $\{Q_j\}$ is nonempty. Consider the cycle

\Gamma_{n}:=\sum_{j\in J}\partial Q_j.\\

Set

U_n:=\operatorname{int}\Bigl(\bigcup_{j\in J}Q_j\Bigr),\qquad\Gamma_n':=\partial U_n,\qquad V_n:=\bigcup_{j\in J}(\operatorname{int}Q_j).\\

Let $\gamma$ be a cycle on $U$ homologous to $0$ . Choose $\delta$ sufficiently small such that $\operatorname{im}(\gamma)\subset U_n$ . Let $\xi\in U\setminus U_n$ , then there exists $Q\in\mathcal Q\setminus\{Q_j\}$ such that $\xi\in Q$ and $Q\notin U$ . Pick $\xi_0\in Q\setminus U$ . Since $\gamma$ is homologous to $0$ , $n(\gamma,\xi_0)=0$ . We can connect $\xi$ and $\xi_0$ by a straight line $L\subset Q$ such that $L\cap U_{n}=\varnothing$ , then $L\subset\mathbb C\setminus\operatorname{im}(\gamma)$ , meaning $\xi_0,\xi$ are in the same connected component of $\mathbb C\setminus\operatorname{im}(\gamma)$ . Since the winding number is continuous in the point, $n(\gamma,\xi)=n(\gamma,\xi_0)=0$ . Since $\Gamma_n'\subset U\setminus U_n$ , for any $\xi\in\Gamma_n'$ , $n(\gamma,\xi)=0$ .

Let $z\in V_n$ , then there exists a unique $j_0\in J$ such that $z\in Q_{j_0}$ . By the Cauchy integral formula for rectangles, we have

\frac{1}{2\pi i}\int_{\Gamma_{n}}\frac{f(w)}{w-z}\,dw=\frac{1}{2\pi i}\int_{\partial Q_{j_0}}\frac{f(w)}{w-z}\,dw=f(z).\\

It is easy to see that the integral along $\Gamma_{n}$ always equals the integral along $\Gamma_n'$ , so

The above equality holds on $V_n$ . Since both sides are continuous, it holds for all $z\in U_n$ . In particular, we have

\int_{\gamma}f\,dz=\int_{\gamma}\left(\frac{1}{2\pi i}\int_{\Gamma_n}\frac{f(w)}{w-z}\,dw\right)\,dz.\\

Since $\Gamma_n'\cap\gamma=\varnothing$ , the integrand is continuous in both variables, so

\begin{aligned} \int_{\gamma}f\,dz&=\int_{\Gamma_n'}\left(\frac{1}{2\pi i}\int_{\gamma}\frac{f(w)}{w-z}\,dz\right)\,dw\\ &=\int_{\Gamma_{n}'}(-n(\gamma,w)f(w))\,dw=0. \end{aligned}\\

If $U$ is unbounded, take $R$ sufficiently large such that $\operatorname{im}(\gamma)\subset D(0,R)$ and set $U':=U\cap D(0,R)$ . $\square$

Corollary 5.5 $\quad$ Let $U\subset\mathbb C$ be a simply connected open set, $\gamma$ a cycle in $U$ , $f$ a holomorphic function on $U$ . Then

\int_{\gamma}f\,dz=0.\\

Theorem 5.6 (General Form of Cauchy Integral Formula) $\quad$ Let $U\subset\mathbb C$ be an open set, $\gamma$ a cycle on $U$ homologous to $0$ . Let $f$ be a holomorphic function on $U$ , $a\in U\setminus\operatorname{im}(\gamma)$ . Then

\frac{1}{2\pi i}\int_{\gamma}\frac{f(z)}{z-a}\,dz=n(\gamma,a)f(a).\\

Proof $\quad$ Set

g(z):=\frac{f(z)-f(a)}{z-a},\\

then $g$ is holomorphic on $U$ , so $\int_{\gamma}g\,dz=0$ . Hence

\frac{1}{2\pi i}\int_{\gamma}\frac{f(z)}{z-a}\,dz=\frac{1}{2\pi i}\int_{\gamma}\frac{f(a)\,dz}{z-a}=n(\gamma,a)f(a).\\

$\square$

Analytic Functions#

Analyticity, Taylor Expansion#

Theorem 6.1 $\quad$ Let $\gamma$ be a rectifiable curve in $\mathbb C$ , $\phi:\operatorname{im}(\gamma)\to\mathbb C$ a continuous function. For $z\in\mathbb C\setminus\operatorname{im}(\gamma)$ , define

f(z):=\int_{\gamma}\frac{\phi(w)}{w-z}\,dw,\\

then for any $z_0\in\mathbb C\setminus\operatorname{im}(\gamma)$ , set $r:=\operatorname{dist}(z_0,\operatorname{im}(\gamma))$ , $f$ can be expanded as a power series in $B(z_0,r)$ :

f(z)=\sum_{n=0}^{\infty}\left(\int_{\gamma}\frac{\phi(w)\,dw}{(w-z_0)^{n+1}}\right)(z-z_0)^n.\\

Proof $\quad$ Take $w\in\operatorname{im}(\gamma)$ , note that

\frac{\phi(w)}{w-z}=\frac{\phi(w)}{w-z_0}\frac{w-z_0}{w-z}=\frac{\phi(w)}{w-z_0}\frac{1}{1-(z-z_0)/(w-z_0)},\\

thus for $\left|\frac{z-z_0}{w-z_0}\right|\leq 1$ , in particular for $z\in B(z_0,r)$ , we have

\frac{\phi(w)}{w-z}=\sum_{n=0}^{\infty}\frac{\phi(w)}{(w-z_0)^{n+1}}(z-z_0)^n.\\

Set $g_n(w):=\sum_{j=0}^{n}\phi(w)(w-z_0)^{-j-1}(z-z_0)^j$ , we will show $g_n\to\phi(w)/(w-z)$ uniformly on $\operatorname{im}(\gamma)$ . Indeed, let $M:=\|\phi\|_u$ and $r':=|z-z_0|<r$ , then

\begin{aligned} \left|g_n(w)-\frac{\phi(w)}{w-z}\right|&=\left|\sum_{j=n}^{\infty}\frac{\phi(w)}{(w-z_0)^{j+1}}(z-z_0)^j\right|\\ &\leq M\sum_{j=n}^{\infty}\left|\frac{(z-z_0)^j}{(w-z_0)^{j+1}}\right|\leq M\sum_{j=n}^{\infty}\left(\frac{r'}{r}\right)^j\frac{1}{r}=\frac{M}{r-r'}\left(\frac{r'}{r}\right)^n, \end{aligned}\\

so $\|g_n-\phi/(w-z)\|_u\to 0$ , proving uniform convergence. Hence

\int_{\gamma}\frac{\phi(w)}{w-z}\,dw=\sum_{n=0}^{\infty}\left(\int_{\gamma}\frac{\phi(w)\,dw}{(w-z_0)^{n+1}}\right)(z-z_0)^n.\\

$\square$

Corollary 6.2 (Taylor Expansion) $\quad$ Let $D:=D(z_0,r)\subset C$ be a disk, $f:\overline D\to\mathbb C$ holomorphic, then $f$ equals a power series on $D$ , in particular

f(z)=\frac{1}{2\pi i}\sum_{n=0}^{\infty}\left(\int_{\partial D}\frac{f(w)\,dw}{(w-z_0)^{n+1}}\right)(z-z_0)^n.\\

Comparing the coefficients from Theorem 6.1 and the Cauchy integral formula with the Taylor series also yields the derivative formula

f^{(n)}(z)=\frac{n!}{2\pi i}\int_{\partial D}\frac{f(\zeta)\,d\zeta}{(\zeta-z)^{n+1}},\\

and this method is simpler than the one we used in Section 4.

Theorem 6.3 (Morera) $\quad$ Let $U\subset\mathbb C$ be an open set, $f:U\to\mathbb C$ continuous. If for every rectifiable closed curve $\gamma$ in $U$ , $\int_{\gamma}f\,dz=0$ , then $f$ is holomorphic.

Proof $\quad$ The condition implies $f\,dz$ is exact. $\square$

Theorem 6.4 $\quad$ Let $U\subset\mathbb C$ be an open set, $\{f_n\}$ a sequence of holomorphic functions on $U$ , converging uniformly to $f$ on every compact subset of $U$ . Then $f$ is also a holomorphic function on $U$ .

Proof $\quad$ For any $\overline D\subset U$ , for $z\in D$ , $f_n(z)=\frac{1}{2\pi i}\int_{\partial D}\frac{f_n(w)}{w-z}\,dw$ , letting $n\to\infty$ gives $f(z)=\frac{1}{2\pi i}\int_{\partial D}\frac{f(w)}{w-z}\,dw$ . By Theorem 6.1, $f$ is holomorphic on $D$ . Thus $f$ is holomorphic on the entire $U$ . $\square$

Cauchy Estimates, Liouville’s Theorem, Fundamental Theorem of Algebra#

Theorem 6.5 (Cauchy’s estimate) $\quad$ Let $D:=B(z_0,r)\subset\mathbb C$ be a disk, $f:\overline{D}\to\mathbb C$ holomorphic. Set $M:=\max_{z\in\partial D}|f(z)|$ . Then

|f^{(n)}(z_0)|\leq Mn!r^{-n}.\\

Proof $\quad$

\begin{aligned} |f^{(n)}(z_0)|&=\left|\frac{n!}{2\pi i}\int_{\partial D}\frac{f(\zeta)\,d\zeta}{(\zeta-z)^{n+1}}\right|\leq\left|\int_{\partial D}Mr^{-n-1}\,ds\right|=Mn!r^{-n}. \end{aligned}\\

$\square$

Theorem 6.6 (Liouville) $\quad$ Let $f:\mathbb C\to\mathbb C$ be a bounded holomorphic function, then $f$ is constant.

Proof $\quad$ Suppose $|f|\leq M$ , let $z_0\in\mathbb C$ , for any $R\in\mathbb R$ apply Cauchy’s estimate on $B(z_0,R)$ to get $|f'(z_0)|\leq MR^{-n}$ , since $R$ is arbitrary, $f'(z_0)=0$ , so $f'\equiv 0$ . Hence $f$ is constant. $\square$

Definition 6.7 $\quad$ If a function $f$ is holomorphic on the entire complex plane, it is called an entire function.

Corollary 6.8 $\quad$ If the real or imaginary part of an entire function $f$ is bounded, then $f$ is constant.

Proof $\quad$ Let $f=u+iv$ . Suppose $u$ is bounded. Consider $g:=\exp(f)$ , then $|g|=|e^u|$ . Since $u$ is bounded, so is $g$ , thus by Liouville’s theorem $g=\exp(f)$ is constant, so $f$ is constant. $\square$

Theorem 6.9 (Fundamental Theorem of Algebra) $\quad$ $\mathbb C$ is algebraically closed: Let $f\in\mathbb C[X]$ , $\deg f\geq 1$ . Then $f$ has at least one root in $\mathbb C$ .

Proof $\quad$ Assume $f$ has no root in $\mathbb C$ , then $1/f$ is holomorphic on all of $\mathbb C$ . Since $\deg\geq 1$ , as $|z|\to\infty$ , $|f|\to\infty$ , so $1/f\to 0$ , hence $1/f$ is bounded. By Liouville’s theorem, $1/f$ is constant, so $f$ is also constant, contradicting $\deg f\geq 1$ . Therefore $f$ has at least one root in $\mathbb C$ . $\square$

Zeros and Uniqueness#

Lemma 6.10 $\quad$ Let $U\subset\mathbb C$ be a connected open set, $f$ a holomorphic function on $U$ , $z_0\in U$ . Suppose for all $n\geq 1$ , $f^{(n)}(z_0)=0$ . Then $f\equiv f(z_0)$ on $U$ .

Proof $\quad$ Set

V:=\{z\in U:f(z)=f(z_0),\,f^{(n)}(z)=0,\,\forall n\geq 1\}.\\

Since $z_0\in V$ , $V$ is nonempty. Clearly $V$ is closed. Let $z\in V$ , then there exists $\overline{D(z,\epsilon)}\subset U$ such that $f$ is holomorphic on $\overline{D(z,\epsilon)}$ , so $f$ can be expanded as a power series in $D$ and all coefficients of nonzero terms are zero, hence $f$ is identically $f(z_0)$ on $D$ , so all derivatives of $f$ are zero on $D$ , thus $D\subset V$ . This shows $V$ is open. Since $U$ is connected, $V=U$ . $\square$

Lemma 6.11 $\quad$ Zeros of holomorphic functions are isolated: Let $U\subset\mathbb C$ be a connected open set, $f$ a non-constant holomorphic function on $U$ , $z_0\in U$ , then there exists $r>0$ such that $z_0$ is the unique zero of $f-f(z_0)$ in $D(z_0,r)$ .

Proof $\quad$ By Lemma 6.10, there exist $r_0>0$ and $m\in\mathbb N_{\geq 1}$ such that on $D':=D(z_0,r_0)$

f(z)=f(z_0)+a_m(z-z_0)^m+\cdots,\quad a_m\neq 0,\\

then on $D'$ , $f-f(z_0)=(z-z_0)^mg(z)$ , where the holomorphic function $g$ satisfies $g(z_0)=a_m\geq 0$ . Since $g$ is continuous, there exists $0<r<r_0$ such that on $D:=D(z_0,r)$ , $g(z)\neq 0$ . Thus $z_0$ is the unique zero of $f-f(z_0)$ on $D$ . $\square$

Corollary 6.12 $\quad$ Let $U\subset\mathbb C$ be a connected open set, $f,g$ holomorphic functions on $U$ . If there exists a sequence of points $\{z_n\}$ such that $z_n\to z\in U$ and $f(z_j)=g(z_j),\quad\forall j\in\mathbb N$ , then $f\equiv g$ on $U$ .

Proof $\quad$ By continuity $f(z_0)=g(z_0)$ , then apply Lemma 6.11. $\square$

Logarithm and Taking Roots#

Theorem 6.13 (Logarithm) $\quad$ Let $U\subset\mathbb C$ be a simply connected open set, $f$ a holomorphic function on $U$ , with $f\neq 0$ pointwise. Then there exists a holomorphic function $g$ on $U$ such that $e^g=f$ .

Proof $\quad$ $f'/f$ is holomorphic on $U$ . For $a\in U$ , choose $\alpha\in\mathbb C$ such that $e^{\alpha}=f(a)$ . For any $z\in U$ , let $\gamma$ be a rectifiable curve connecting $a$ and $z$ , define

g(z):=\alpha+\int_{\gamma}\frac{f'}{f}\,dz,\\

then $g$ is independent of the choice of $\gamma$ , so $g$ is holomorphic and $g'=f'/f$ . Set $h:=\exp(-g)f$ , then $h(a)=1$ . Further

h'(z)=-e^{-g(z)}g'(z)f(z)+f'(z)e^{-g(z)}=0,\qquad\forall z\in U.\\

Thus $h\equiv 1$ . Therefore $e^g=f$ on $U$ . $\square$

Theorem 6.14 (Taking Roots) $\quad$ Let $U\subset\mathbb C$ be a simply connected open set, $f$ a holomorphic function on $U$ , with $f\neq 0$ pointwise. Then there exists a holomorphic function $g$ on $U$ such that $g^n=f$ .

Proof $\quad$ There exists a holomorphic function $h$ such that $e^h=f$ . Set $g:=e^{h/n}$ , then $g^n=f$ . $\square$

Open Mapping Theorem, Biholomorphic Mappings, Local Normal Form#

Definition 6.15 (Biholomorphic function) $\quad$ Let $U,V$ be two open sets in $\mathbb C$ . A holomorphic function $f:U\to V$ is called biholomorphic if $f$ is bijective and $f^{-1}V\to U$ is holomorphic.

By the inverse function theorem, if $f:U\to\mathbb C$ is holomorphic, $z_0\in U$ , and $f'(z_0)\neq 0$ , then $f$ is locally biholomorphic at $z_0$ .

Theorem 6.16 (Local Normal Form of Holomorphic Functions) $\quad$ Let $f$ be holomorphic on an open set $U\subset\mathbb C$ , $z_0\in U$ , $f^{(m)}(z_0)\neq 0$ and $\forall j\leq m-1,\,f^{(j)}(z_0)=0$ . Then there exist $D:=D(z_0,r)\subset U$ and a biholomorphic function $h$ on $D$ such that on $D$ , $f-f(z_0)=h^m$ .

Proof $\quad$ Locally on $D':=D'(z_0,r)$ we can write $f-f(z_0)=(z-z_0)^mg(z)$ , where $g$ is holomorphic, $g(z_0)=a_m\neq 0$ . Let $\tilde h$ be a holomorphic function on $D'$ satisfying $\tilde h^m=g$ , $h:=(z-z_0)\tilde h$ , then $h$ is holomorphic on $D'$ and $f-f(z_0)=h^m$ . Note that $h'(z_0)=\tilde h(z_0)\neq 0$ , so by the inverse function theorem, there exists $D\subset D'$ such that $h$ is biholomorphic on $D$ . $\square$

Lemma 6.17 $\quad$ $g_m(z):\mathbb C\to\mathbb C\,,z\mapsto z^m$ is a proper and open map; further, if $z\neq 0$ , then $g_m^{-1}(z)$ contains exactly $m$ points.

Proof $\quad$ It is easy to see $g_m(z)$ is proper. Let $z=re^{i\theta}$ , consider the equation $w^n=z$ . Let $w=\rho e^{i\psi}$ , then $\rho^m=r$ , $m\psi=\theta+2k\pi$ . The real number $\rho=r^{1/m}$ is unique, while $\phi$ has exactly $m$ solutions modulo $2\pi$ . This proves $g_m^{-1}(z)$ has $m$ points. $\square$

Theorem 6.18 (Open Mapping Theorem) $\quad$ Let $f$ be a non-constant holomorphic function on a connected open set $U\subset\mathbb C$ , then $f$ is an open map.

Proof $\quad$ It suffices to show that for any $z_0\in U$ , there exists $r>0$ such that for any $r'<r$ , $D'=D(z_0,r')$ , $f(D')$ is open. Indeed, by Theorem 6.14, there exist $m\geq 1$ , a biholomorphic function $h$ , and a disk $D=:D(z_0,r)$ such that on $D$ , $f-f(z_0)=h^m$ . By Lemma 6.15, $z\mapsto z^m$ is open, so such $D$ is as desired. $\square$

Theorem 6.19 $\quad$ Let $f$ be a holomorphic function on $\mathbb U$ . If $f$ is injective, then $f:U\to f(U)$ is biholomorphic.

Proof $\quad$ It suffices to prove the proposition for each connected component of $U$ . Below assume $U$ is connected. $f(U)$ is open. $\forall z_0\in U$ , $f'(z_0)\neq 0$ , otherwise by Theorem 6.14, $f$ is at least $m\,(m\geq 2)$ -to-one around $z_0$ , so by the inverse function theorem, $f^{-1}$ is locally holomorphic at $f(z_0)$ . Since $z_0$ is arbitrary, $f$ is biholomorphic. $\square$

Maximum Modulus Principle#

Theorem 6.20 (Maximum Modulus Principle) $\quad$ Let $U\subset\mathbb C$ be a connected open set, $f$ a non-constant holomorphic function on $U$ . Then there does not exist $z_0\in U$ such that $|f(z_0)|=\sup_{z\in U}|f(z)|$ .

Proof $\quad$ This is because $f(U)$ is open. $\square$

Considering neighborhoods $U'\subset U$ for any $z_0\in U$ in the maximum modulus principle, it further follows that $f$ cannot attain a local maximum at any $z_0\in U$ . Similarly, we have the dual “minimum modulus principle”.

Proposition 6.21 $\quad$ Let $U\subset\mathbb C$ be a connected open set, $f$ a non-constant holomorphic function on $U$ , with $f\neq 0$ pointwise. Then there does not exist $z_0\in U$ such that $|f(z_0)|=\inf_{z\in U}|f(z)|$ .

Corollary 6.22 $\quad$ Let $U\subset\mathbb C$ be a bounded connected open set, $f$ a non-constant holomorphic function on $\overline U$ . Then the maximum of $|f|$ on $\overline U$ is attained on $\partial U$ . If further $f$ is nonvanishing pointwise, then the minimum of $|f|$ is attained on $\partial U$ .

Laurent Series#

Let $z_0\in\mathbb C$ . A Laurent series centered at $z_0$ is a series of the form:

\sum_{n=1}^{\infty}b_n(z-z_0)^{-n}+\sum_{n=0}^{\infty}a_n(z-z_0)^n,

where $a_n,b_n\in\mathbb C$ . $\sum_{n=1}^{\infty}b_n(z-z_0)^{-n}$ is called the principal part, $\sum_{n=0}^{\infty}a_n(z-z_0)^n$ the regular part.

Let $f:=\sum_{n=1}^{\infty}b_n(z-z_0)^{-n}+\sum_{n=0}^{\infty}a_n(z-z_0)^n$ , $R$ the radius of convergence of its regular part. For the principal part, we can also define a “radius of divergence” $r$ . Indeed, set $S:=\sum_{j=0}^{\infty}b_jz^j$ , consider the radius of convergence $R'$ of $S$ , then the principal part converges if and only if $S\big((z-z_0)^{-1}\big)$ converges. Thus if $|(z-z_0)^{-1}|>R'$ i.e., $|z-z_0|<1/R'$ , the principal part does not converge, conversely, if $|z-z_0|>1/R'$ , the principal part converges. So we can set $r=1/R'$ . Define the open annulus

D(z_0,r,R):=\{z\in\mathbb C:r<|z-z_0|<R\},

then $f$ converges uniformly on compact subsets of $D(z_0,r,R)$ . Since each partial sum is holomorphic, $f$ is holomorphic on $D(z_0,r,R)$ .

Lemma 7.1 $\quad$ Let $g$ be a holomorphic function on $D(z_0,r,R)$ , $r<r_1<r_2<R$ . Then

\int_{\partial D(z_0,r_1)}g(z)\,dz=\int_{\partial D(z_0,r_2)}g(z)\,dz.

Proof $\quad$ Set $\gamma:=\partial D(z_0,r_1)-\partial D(z_0,r_2)$ . Since $\partial D(z_0,r_j)$ are homotopic, $\gamma$ is homologous to $0$ in $D(z_0,r,R)$ , so by the general form of Cauchy’s integral theorem, $\int_{\gamma}g\,dz=0$ , proving the lemma. $\square$

Theorem 7.2 $\quad$ Let $f=\sum_{n=-\infty}^{\infty}a_n(z-z_0)^n$ be a Laurent series convergent on $D(z_0,r,R)$ , $r<\rho<R$ . Then

a_n=\frac{1}{2\pi i}\int_{\partial D(z_0,\rho)}\frac{f(w)}{(w-z_0)^{n+1}}\,dw.

Proof $\quad$ By uniform convergence,

\int_{\partial D(z_0,\rho)}\frac{f(w)}{(w-z_0)^{n+1}}\,dw=\sum_{k=-\infty}^{\infty}\int_{\partial D(z_0,\rho)}\frac{a_k\,dw}{(w-z_0)^{n+1-k}}.

If $k\neq n$ then $a_k(w-z_0)^{k-n-1}$ is exact, its integral is zero. So

\begin{aligned} \int_{\partial D(z_0,\rho)}\frac{f(w)}{(w-z_0)^{n+1}}\,dw=\int_{\partial D(z_0,\rho)}\frac{a_n\,dw}{w-z_0}=2\pi i\cdot a_n. \end{aligned}

$\square$

Theorem 7.3 (Laurent Expansion) $\quad$ Let $f$ be a holomorphic function on $D(z_0,r,R)$ , then $f$ can be represented as a Laurent series centered at $z_0$ on $D(z_0,r,R)$ .

Proof $\quad$ By Theorem 7.2, the coefficients of the Laurent series, if they exist, are unique. It suffices to show that $f$ can be expanded as a Laurent series on each $D(z_0,r',R'),\,r<r'<R'<R$ . For each $z\in D(z_0,r',R')$ , $n(\partial D(z_0,R')-\partial D(z_0,r'),z)\equiv 1$ . Then by the general form of Cauchy’s integral theorem,

f(z)=\frac{1}{2\pi i}\int_{\partial D(z_0,R')}\frac{f(w)}{w-z}\,dw-\frac{1}{2\pi i}\int_{\partial D(z_0,r')}\frac{f(w)}{w-z}\,dw,

for fixed $z\in D(z_0,r',R')$ , consider the series

\frac{w-z_0}{w-z}=\sum_{n=0}^{\infty}\left(\frac{z-z_0}{w-z_0}\right)^n,

it converges for $w\in D(z_0,R')$ , so

\int_{\partial D(z_0,R')}\frac{f(w)}{w-z}\,dw=\sum_{n=0}^{\infty}\left(\int_{\partial D(z_0,R')}\frac{f(w)\,dw}{(w-z_0)^{n+1}}\right)(z-z_0)^n.

For fixed $z\in D(z_0,r',R')$ , consider the series

\frac{z-z_0}{z-w}=\sum_{n=0}^{\infty}\left(\frac{w-z_0}{z-z_0}\right)^n,

it converges for $w\in D(z_0,r')$ , so

\begin{aligned} \int_{\partial D(z_0,r')}\frac{f(w)}{w-z}\,dw&=-\sum_{n=0}^{\infty}\left(\int_{\partial D(z_0,r')}(w-z_0)^nf(w)\,dw\right)(z-z_0)^{-n-1}\\ &=-\sum_{n=-1}^{-\infty}\left(\int_{\partial D(z_0,r')}\frac{f(w)\,dw}{(w-z_0)^{n+1}}\right)(z-z_0)^n. \end{aligned}

Combining these two parts proves the proposition. $\square$

From Theorem 7.2 and Theorem 7.3, if $f$ is holomorphic on $D(z_0,r,R)$ , then $f$ can be expanded as

f(z)=\frac{1}{2\pi i}\sum_{n=-\infty}^{\infty}\left(\int_{\partial D(z_0,\rho)}\frac{f(\zeta)\,d\zeta}{(\zeta-z_0)^{n+1}}\right)(z-z_0)^n.

Analytic Functions and Power Series Theory

https://peakyi.github.io/posts/analytic-functions-and-power-series-theory/

Author

Anby Demara

Published at

2025-12-10

License

CC BY-NC-SA 4.0

Note on Complex Analysis

NOIP2025